Mass between Two Damped Springs

initial position (m)

-2

mass (kg)

0.8

spring constant 1 (N/m)

2

spring constant 2 (N/m)

1.7

viscous damping coefficient γ

N·s

m

0

time

20

Consider a mass

m

between two springs attached to opposing walls. Let

k

1

and

k

2

be the respective spring constants. A displacement of the mass by a distance

x

results in the first spring lengthening by

x

, while the second spring is compressed by

x

(and pushes in the same direction). Suppose that the equilibrium position is at

x=0

, where the two springs have their force-free lengths. The equation of motion is then given by

2

d

x

d

2

t

+

γ

m

dx

d t

+

2

ω

0

x=0

, where

γ

is the viscous damping coefficient in

Ns/m

and

ω

0

=

k

1

+

k

2

m

is the natural frequency of the system. Evidently, the two springs are dynamically equivalent to a single spring with constant

k=

k

1

+

k

2

. The solution is

x(t)=A

-γt/2m

e

sin

1

2m

4

2

m

2

ω

0

-

2

γ

t

. When

γ=0

(no damping), this reduces to

x(t)=Asin(

ω

0

t)

. The motion is traced by the red curves in the right panel.